如何添加数组的偶数和奇数的和 [英] how to add a sum of even and odd number of an array

问题描述

我想从一个递归数组中获得偶数的和，然后将一个奇数的和加在一起。我试图循环通过数组，以便得到偶数，但它保持返回第一个索引。请帮助和提前感谢。

i am trying to get the sum of even numbers from a recursive array and the sum of an odd number and then add both sums together. i am trying to loop through the arrays so that to get the even numbers but it keep return the first index. please help and thanks in advance..

我的目标是取数组2,1,5,9,8,4取偶数索引，并将其添加到奇数索引。 a [0] = 2，a [1] = 1，a [2] = 5，a [3] = 9，a [4] = 8，所以它需要（2 + 5 + 8） - （1 + 9 + 4）= 1

my aim is to take in array 2,1,5,9,8,4 take the even index and add it to the odd indexes. a[0]=2,a[1]=1,a[2]=5, a[3]=9, a[4]=8, a[5]=4. so it would take (2+5+8)-(1+9+4)=1

这是我到目前为止我不熟悉递归我的代码可能已关闭

this is what i got so far i am not familiar with recursive so my code might be off

int calc(int *a, int size)
{
    if(size==1 || size==0)
        return a[0];
    for(int i=0; i<size; i++){
        if(i%2==0){
            int sum_i = a[i];
            int m=calc(a, size-1);
            if(m>a[size-1])
                return m;
        }
    }
    for(int j=0; j<size; j++){
        if(j%2!=0);
        int sum_j = a[j];
        return sum_j;
    }
    int sum = a[i] - a[j];
    int e = calc(a, size-1);
    if(e%2==0)
        return e=e+0; //return even

    return sum;
}

int main( )
{
    int a[6]={1,2,3,5,6,2};
    int size = 6;   
    cout<< calc(a, size)<<endl;

    system("pause");   
    return 0;
}

推荐答案

// calc(): returns sum(a[0], a[2], a[4], ...) - sum(a[1], a[3], a[5], ...)
int calc(int *a, int size)
{
    int sum_even_pos = 0;
    int sum_odd_pos = 0;
    for (int i = 0; i < size; i++) 
    {
        sum_even_pos += a[i];
        if (++i < size) sum_odd_pos += a[i];
    }
    return sum_even_pos - sum_odd_pos;
}

http://codepad.org/DVx0sgg1

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